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Let r cm and h cm be the radius and the height of the cone. Then,

h =$\frac{1}{6}$ r. H = 6h

So, volume, V of the cone =$\frac{1}{3}$ πr²h

$=\frac{1}{3}\pi {(6h)}^2\times h.$ $\frac{4d}{dt}.$

= 12 × h³

Rate of change of volume of the cone wrt the height is

$\frac{\mathrm{dV}}{dh}=\frac{d}{dh}\left(12\pi h^3\right)$ = 12 × π × 3 × h².

As the sand is pouring from the pipe at rate of 12${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

we have

$\frac{d}{dt}=12$

$\Rightarrow \frac{dv}{dh}\times \frac{dh}{dt}=12$

$\Rightarrow 36\pi h^2\frac{dh}{dt}=12.$

$\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi h^2}=\frac{1}{3\pi h^2}.$

${\therefore \frac{dh}{dt}|}_{h=4 }=\frac{1}{3\pi \times {(4)}^2}=\frac{1}{48\pi }.$

Hence, the height is increasing at the rate of$\frac{1}{48x}$ cm/s.

Given, diameter of the spherical balloon = $\frac{3}{2}$ (2x + 1)

So, radius of the spherical r = $\frac{1}{2}\times \frac{3}{2}(2x+1)$

$=\frac{3}{4}(2x+1)$

Then, volume of the spherical V = $\frac{4}{9}\pi r^3$

$=\frac{4}{3}\pi \times [\frac{3}{4}{\left(\frac{3}{(2x+1)}\right]}^3$

$=\frac{9\pi }{16}{(2x+1)}^3.$

Q Rate of change of volume wrt.tox, $\frac{dV}{dx}=\frac{d}{dx}\left[\frac{\pi }{16}{(2x+1)}^3\right]$

$=\frac{9\pi }{16}\times 3\times {(2x+1)}^2·\frac{d}{dx}(2x+1)$

$=\frac{27\pi }{16}{(2x+1)}^2\times 2=\frac{27\pi }{8}{(2x+1)}^2.$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv) is the correct answer

The strongest oxidising agent means it has greater tendency to oxidise other species and itself gets easily reduced. So higher the E^? values, stronger is the oxidising agent it is. Thus, Ag⁺ having the highest positive E^?   value among the given systems, is the strongest oxidising agent.

Let x be the radius of the bubble with volume .V. then,

$\frac{dr}{dt}=\frac{1}{2}$ cm/s

andV = $\frac{4}{3}\pi n^3$

Rate of change of volume $\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}{\pi }^3\right)$ = $\frac{4}{3}\pi \frac{d}{d{t}^4}{r}^3$

$=\frac{4}{3}\pi \times 3r^2\frac{dr}{dt}.$

= 4πr² ×$\frac{1}{2}$

= 2πr².

${\frac{dv}{dt}|}_{r=cm}$ = 2x (1)² 2π. ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

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Given eqⁿ of the curve is 6y = x³ + 2.______ (1)

Wheny coordinate change s 8 times as fast as x-coordinate

$\frac{dy}{dx}$ = 8 _____ (2)

Now, differentiating eqⁿ (1) wrt.x we get,

$\frac{d}{dx}(6y)=\frac{d}{dx}\left(x^3+2\right)$

$\Rightarrow 6\frac{dy}{dx}=3x^2+0.$

6 × 8 = 3x² (using eqⁿ (2)

$\Rightarrow x^2=\frac{48}{3}=16$

$\Rightarrow +\surd x=\pm \surd 16$

x = $\pm$4.

When x = 4, we have, 6y = 4³+ 2 = 64 + 2 + 66

$\Rightarrow y=\frac{66}{6}$ y =11.

And when x = -4, we have, 6y = ( -4)3 + 2 = -64 + 2 = -62

$\Rightarrow y=\frac{62}{6}=\frac{31}{3}$

The tequired point s are (4, 11) and $\left(-4,\frac{31}{3}\right)$

Since, the bottom of ground is increasing with time t,

$\frac{dx}{dt}$ = 2cm/s

From fig, Δ ABC, by Pythagorastheorem

AB² + BC² = AC²

$\Rightarrow$ x² + y² = 5²

x² + y² = 25 ____ (1)

Differentiating eqⁿ (1) w. r. t. time t we get,

$\frac{d}{dt}\left(x^2+y^2\right)=\frac{d}{dt}(25)$

$\Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.$

$\Rightarrow 2x\times 2+2y\frac{2y}{dy}=0$

$\Rightarrow \frac{dy}{dt}=-\frac{2x}{y}$ m/s

When x = 4m, the rate at which its height on the wall decreases is

$\frac{dy}{dt}=-\frac{2\times 4}{3}$ $\{\begin{array}{l}\therefore 4^2+y^2=5^2\\ \Rightarrow y^2=25-16\\ \Rightarrow y\text{\hspace{0.17em}√9}\left(L\text{\hspace{0.17em}engthcan'tbenegetive}\right)\\ \Rightarrow y=3\end{array}$

$\Rightarrow \frac{dy}{dt}=-\frac{8}{3}$ room

The volume v of a spherical balloon with radius r is V.$\frac{4}{3}\pi r^3.$

with respect its radius.

Then, the rate of change of volume $|\frac{\mathrm{dV}}{dr}=\frac{d}{dr}\left(\frac{4}{3}·\pi r^3\right)$

$=\frac{4}{3}\pi \left(\frac{d}{dr}{r}^3\right)$

$=\frac{4}{3}\pi \times 3\times {\pi }^2$

= 4$\pi$ r²

Whenx = 10 cm,

$\frac{\mathrm{dV}}{dr}$ = 4 $\pi$10)² = 400 $\pi$cm³/cm

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Let ‘r’ cm be the radius of volume V. measured Then,

$V\; =\frac{4}{3}\pi r^3$

Now, rate at which balloon is being inflated = 900${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{dv}{dr}=900$ ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$ = 900 ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dr}\left(\frac{4}{3}\times r^3\right)\frac{dr}{dt}=900$

$\frac{4}{3}$$\pi$ × 3 × r² $\frac{dr}{dr}$ = 900.

$\Rightarrow \frac{dr}{dt}=\frac{900}{4\pi r^2}.$

When r = 15cm,

${\frac{ds}{dt}|}_{r=15}=\frac{900}{4\pi \times {(15)}^2}$ = $\frac{900}{900\pi }=\frac{1}{\pi }$ cm/s.

Q Radius of balloon increases by $\frac{1}{\pi }$ per second.

This is a Short answer type question as classified in NCERT Exemplar

(i) The balanced chemical is given as –

 

(ii) The balanced chemical is given as –

 

(iii) The balanced chemical is given as –

 

(iv) The balanced chemical is given as –

 

 

 

 

Since the length x is decreasing and the widthy is increasing with respect to time we have

$\frac{dx}{dt}$ = 5cm/min and $\frac{dy}{dt}$ = A cm/min

(a) The perimeter P of a rectangle will be, P = 2 (x + y)

Q  Rate of change of perimeter, $\frac{dP}{dt}=\frac{d^2}{dt}(x+y)$

$=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)$

= 2 ( -5 + 4)

= -2 cm/min

(b) The area A of the rectangle is A = x. y.

Rate of change of area is $\frac{\mathrm{dA}}{dt}=\frac{d}{dt}(x·y)$

$=x\frac{dy}{dt}+\frac{dx}{dt}·y.$

= 4x - 5y

So,  $\frac{\mathrm{dA}}{dt}$ |x = 8ay = 6cm = 4 (8) - 5 (6) = 32 - 30 = 2 cm²/min spherical balloon

The circumference C of the circle with radius r is C = 2πr.

Then, rate of change in circumference is $\frac{\mathrm{dC}}{dt}=\frac{d}{dt}2\pi r$ = $2\pi \frac{dx}{df}.$

Q Radius of circle increases at rate 0.7 cm/s we get,

$\frac{dy}{dt}=0.7$ cm/s

So,  $\frac{\mathrm{dC}}{dt}$ = 2.× 0.7 cm/s = 1.4 × cm/s.

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The area A of the circle with radius π is A = πr²

Then, rate of change in area $\frac{\mathrm{dA}}{dt}=\frac{d\stackrel{?}{x}{r}^2}{dt}$ = $\frac{dx{r}^2}{dr}·\frac{dr}{dt}$

= 2πr $\mathrm{dr}$
${\frac{}{dt}}_{.}$

Q The wave moves at a rate 5cm/s we have,

$\frac{dr}{\mathrm{dF}}$ = 5cm/s

So,  $\frac{\mathrm{dA}}{dt}$ =r. 5 = 10πr. ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When r = 8 cm.

$\frac{d}{dt}$ = 10.π.8 cm ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ = 80 × cm ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

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Let 'x' cm be the length of edge of the cube which is a fxⁿ of time t then,

$\frac{dx}{dt}$ = 3cm/s as it is increasing.

Now, volume v of the cube is v = x³

Ø Rate of change of volume of the cube $\frac{dv}{dt}=\frac{d{q}^3}{dt}.$

$=\frac{d{x}^3}{dx}\frac{dx}{dt}$

= 3x².3

= 9x² ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When x = 10cm.

$\frac{dv}{dt}$ = 9 x (10)²= 900 ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

Let ‘r’ cm be the radius of the circle which is afxn of time.

Then,  $\frac{dy}{dt}$ = 8 3cm/s as it is increasing.

Now, the area A of the circle is A = πr2.

So, the rate at which the area of the circle change $\frac{d}{dt}=\frac{d}{dt}$ πr2.

$=\frac{d}{dr}\pi r^2·\frac{dr}{dt}$

= 2πr 3

= 6πr. ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When r = 10cm,

$\frac{d}{dt}$ = 6.π × 10 = 60π ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$